Question

Find the antilog of the number $\left( 3.9333 \right)$

Answer 1

Hint: To solve this question first of all we will be using the formula for logarithm. Secondly we will use value for log using the log table. Lastly after using the formulas and values, we can easily find the value which is asked in the given question.

Complete step-by-step solution -
In most of the exams, a log or antilog table is not provided to us. So in this solution we will understand how to find the value for log or antilog without using a log table or antilog table.
The mostly used log formulas are
(a)${{\log }_{e}}=2.303{{\log }_{10}}$
We mostly use ‘log’ as base 10 i.e. ${{\log }_{10}}$
(b) $\log \left( a\times b \right)=\log a+\log b$
(c) $\log \left( \dfrac{a}{b} \right)=\log a-\log b$
(d) $\log {{a}^{b}}=b\left( \log a \right)$
Let us write some values for ‘log’ using log table
$\begin{align}
  & {{\log }_{10}}1=0 \\
 & {{\log }_{10}}2=0.3 \\
 & {{\log }_{10}}3=0.47 \\
 & {{\log }_{10}}4=0.6 \\
 & {{\log }_{10}}5=0.7 \\
 & {{\log }_{0}}6=0.77 \\
 & {{\log }_{10}}7=0.85 \\
 & {{\log }_{10}}8=0.93 \\
 & {{\log }_{10}}9=0.95 \\
 & {{\log }_{10}}10=1 \\
\end{align}$
Now will be finding the value of antilog $\left( 3.9333 \right)$.
$\begin{align}
  & =anti\log \left( 3.9333 \right) \\
 & =anti\log \left( 3+0.9333 \right) \\
 & =anti\log \left( \log {{10}^{3}}+\log 8 \right) \\
\end{align}$
For the first term we will use the formula $\log {{a}^{b}}=b\left( \log a \right)$ and for the second term we will substitute the value for ${{\log }_{10}}8=0.93$
\[\begin{align}
  & =anti\log \left\{ \log \left( {{10}^{3}}\times 8 \right) \right\} \\
 & =anti\log \times \log \left( {{10}^{3}}\times 8 \right) \\
 & =\dfrac{1}{\log }\times \log \left( {{10}^{3}}\times 8 \right) \\
 & =8\times {{10}^{3}} \\
\end{align}\]
Hence the value of $anti\log \left( 3.9333 \right)$ is $8\times {{10}^{3}}$.

Note: From the above solution we conclude that we can easily find the value for ‘log’ or ‘antilog’ without using the ‘log’ or ‘antilog’ table. We can also find the negative value of an antilog by using the same method. Always keep in mind a good command over the logarithm formula as well as the values for log from 1-10 must be required before solving this kind of question.
Example:
$\begin{align}
  & =anti\log \left( -5.7 \right) \\
 & =anti\log \left( -6+0.3 \right) \\
 & =antiog\left( {{\log }_{10}}{{10}^{-6}}+\log 2 \right) \\
\end{align}$
In the above for the first term we have use the formula $\log {{a}^{b}}=b\left( \log a \right)$ and for the second term we have just use the value for ${{\log }_{10}}2=0.3$
\[\begin{align}
  & =anti\log \left\{ \log \left( {{10}^{-6}}\times 2 \right) \right\} \\
 & =anti\log \times \log \left( {{10}^{-6}}\times 2 \right) \\
 & =\dfrac{1}{\log }\times \log \left( {{10}^{-6}}\times 2 \right) \\
 & =2\times {{10}^{-6}} \\
\end{align}\]
Thus we can find the antilog of negative value too.